Q.Derive Bragg's equation.

Bragg's equation, nλ = 2d sinθ, is derived by finding the path-difference condition for constructive interference between X-rays reflected from two successive parallel planes of atoms in a crystal.

Setup: Consider a crystal made up of a set of parallel planes of atoms/ions, separated by a fixed interplanar spacing d. A beam of monochromatic X-rays of wavelength λ falls on these planes at a glancing angle θ (the angle between the incident ray and the plane, not the normal).

Each atom in a plane scatters (reflects) the X-rays. Consider two parallel rays: one reflecting off the top plane, and one penetrating to and reflecting off the very next plane below it (separated by distance d).

Path difference: Using simple geometry, if a perpendicular is dropped from an atom on the upper plane onto the incident and reflected rays of the lower plane, the extra distance travelled by the ray reflected from the lower plane (compared to the ray reflected from the upper plane) is:

Path difference = 2d sinθ

(This comes from the two right-triangles formed at the point of reflection on the lower plane: each contributes an extra path length d sinθ, one on the incident side and one on the reflected side, giving 2d sinθ in total.)

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Bragg's equation, nλ = 2d sinθ, comes from requiring the path difference between X-rays reflected off successive crystal planes to be a whole number of wavelengths.

Derivation of Bragg's Equation:

Consider a crystal made up of a set of parallel planes of atoms/ions, separated by a constant interplanar spacing $d$. A beam of monochromatic X-rays of wavelength $\lambda$ is incident on these planes at a glancing angle $\theta$.

Let two parallel rays 1 and 2 strike two successive planes (upper plane and the plane just below it, separated by distance $d$) at points A and B respectively (with AB perpendicular to the planes). Both rays are reflected at the same angle $\theta$ (angle of incidence = angle of reflection, as in ordinary reflection).

Draw perpendiculars from A to the reflected ray from B (at point C) and to the incident ray line touching B (at point M), such that AM and AC represent the extra path travelled by ray 2 compared to ray 1.

From the geometry of the right triangles formed:

$MB = AB \sin\theta = d\sin\theta$

$BC = AB \sin\theta = d\sin\theta$

So the total extra path length travelled by ray 2 (relative to ray 1) is:

$MB + BC = 2d\sin\theta$

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Bragg's equation nλ = 2d sinθ is derived from the condition that X-rays reflected from successive crystal planes interfere constructively.

When a beam of X-rays of wavelength λ strikes a crystal at a glancing angle θ, it is reflected from parallel sets of atomic planes separated by an interplanar spacing d. Constructive interference (a strong diffracted beam) occurs only when the path difference between rays reflected from successive planes is a whole number multiple of the wavelength.

Derivation:

Consider two parallel planes of atoms, separated by distance d. Two parallel X-ray beams strike these planes at glancing angle θ. Let the rays be reflected from points on the first and second planes.

Draw perpendiculars from a point on the upper plane to the incident and reflected rays that hit the lower plane. Using simple trigonometry, the extra path travelled by the ray reflected from the lower (second) plane compared with the ray reflected from the upper (first) plane is:

Path difference = 2d sinθ

(each of the two perpendicular segments contributes d sinθ, by the right-angled geometry formed with the glancing angle θ)

For constructive interference (i.e., for the reflected beams to reinforce and give a detectable diffracted beam), this path difference must equal a whole number (n = 1, 2, 3, ...) of wavelengths:

nλ = 2d sinθ

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Bragg's law, nλ = 2d sinθ, gives the condition for constructive interference of X-rays diffracted by crystal planes, and is used to determine interplanar spacing.

Derivation of Bragg's equation:

Consider a crystal made up of a set of parallel planes separated by an interplanar distance $d$. A monochromatic beam of X-rays of wavelength $\lambda$ is incident on these planes at a glancing angle $\theta$.

Consider two parallel rays, one reflected from the top plane (at point A) and another from the plane immediately below it (at point B, directly below A by the perpendicular distance $d$). The second ray travels an extra path length compared to the first ray, before and after reflection.

From the geometry (drop perpendiculars from A to the incident and reflected rays of the lower-plane ray), the extra path length travelled by the ray reflected from the second plane is:

Extra path = PB + BQ = $d\sin\theta + d\sin\theta$ = $2d\sin\theta$

(where P and Q are the feet of the perpendiculars from the reflection point on the upper plane onto the incident and reflected rays of the lower-plane reflection.)

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Bragg's equation gives the condition for constructive interference of X-rays reflected from successive planes in a crystal, allowing interplanar spacing to be measured.

Derivation: Consider a crystal made of a set of parallel planes of atoms separated by an interplanar spacing $d$. A monochromatic beam of X-rays of wavelength $\lambda$ strikes these planes at a glancing angle $\theta$ and is partially reflected from each plane (angle of incidence = angle of reflection, as in ordinary reflection).

Consider two parallel rays reflected from two successive planes, ray 1 from the upper plane and ray 2 from the plane immediately below it (separated by distance $d$). Ray 2 travels an extra path compared with ray 1, equal to the distance $BC + CD$, where $B$ and $D$ are the points at which the perpendiculars from the upper plane meet ray 2's path on the lower plane.

From the geometry of the right triangles formed by the incident/reflected rays and the perpendicular to the planes:

$BC = CD = d\sin\theta$

so the total extra path travelled by ray 2 is:

$\text{Path difference} = BC + CD = 2d\sin\theta$

For the two reflected rays to interfere constructively (reinforce each other, giving a diffraction maximum), this path difference must equal a whole number of wavelengths:

$2d\sin\theta = n\lambda \qquad (n = 1, 2, 3, \dots)$

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Bragg's equation, n(lambda) = 2d sin(theta), gives the condition for constructive interference of X-rays diffracted by the parallel planes of atoms in a crystal, and is derived from the geometric path difference between rays reflected off successive planes.

Consider a beam of monochromatic X-rays of wavelength lambda incident at a glancing angle theta on a set of parallel crystal planes separated by an interplanar spacing d. Part of the beam is reflected from the first (upper) plane, and part penetrates to be reflected from the second (lower, parallel) plane, and so on.

Let the rays PA and P'A'B' be two parallel incident rays striking the first and second planes respectively at A and B, and being reflected as AC and BD. Drop perpendiculars from A onto the ray hitting the second plane, meeting it at points E and F, so that AE and AF are perpendicular to the incident and reflected rays through B.

The extra distance travelled by the ray reflected from the second (lower) plane compared with the ray reflected from the first (upper) plane is the path difference:

Path difference = EB + BF

Since the angle of incidence equals the angle of reflection (both = theta, measured from the plane), and AB = d (the interplanar spacing), simple trigonometry on the right triangles AEB and AFB gives:

EB = d sin(theta) and BF = d sin(theta)

So the total path difference = 2d sin(theta).

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Bragg's equation, nλ = 2d sinθ, is derived by considering the path difference between X-rays reflected from two successive parallel planes of a crystal.

Setup

Consider a beam of monochromatic X-rays of wavelength λ striking a crystal at a glancing angle θ, and being reflected from two successive parallel planes of atoms in the crystal, separated by an interplanar spacing d. Let the two rays be reflected from points A (on the first plane) and B (on the second plane).

Path difference

Draw perpendiculars from A onto the incident and reflected rays striking B, meeting them at points P and Q respectively. The extra distance travelled by the ray reflected from the second (lower) plane, compared to the ray reflected from the first plane, is the path difference:

$\text{Path difference} = PB + BQ$

From the geometry of the right triangles APB and AQB (with AB = d, and the angle between AB and the planes = θ):

$PB = d\sin\theta \quad \text{and} \quad BQ = d\sin\theta$

So the total path difference is:

$PB + BQ = 2d\sin\theta$

Condition for constructive interference …

Bragg's equation, nλ = 2d sinθ, gives the condition for constructive interference of X-rays diffracted from parallel planes of atoms in a crystal, and is derived from the path difference between rays reflected from successive planes.

Setup: Consider a beam of monochromatic X-rays of wavelength $\lambda$ incident on a crystal at a glancing angle $\theta$ to a set of parallel atomic planes separated by an interplanar spacing $d$. Part of the beam is reflected from the first plane, and part penetrates to be reflected from the second (and deeper) parallel planes.

Derivation: Consider two parallel rays, Ray 1 reflecting off the top plane at point A, and Ray 2 penetrating to the second plane and reflecting at point C (directly below A), both at the same glancing angle $\theta$ (angle of incidence = angle of reflection, as in ordinary reflection).

Drop perpendiculars from A onto the incident and reflected rays of Ray 2, meeting them at points B and D respectively. The extra path length travelled by Ray 2 compared to Ray 1 is:

$\text{Path difference} = BC + CD$

Since A, B, C, D form right triangles with the interplanar spacing $d = AC$, and the glancing angle is $\theta$:

$BC = CD = d\sin\theta$

So the total path difference is:

$\text{Path difference} = 2d\sin\theta$

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